Let $\alpha=\sqrt[5]{2}+\sqrt{5}$, show that $\sqrt{5}\in\mathbb{Q}(\alpha)$.

Question

Let $\alpha=\sqrt[5]{2}+\sqrt{5}$, show that $\sqrt{5}\in\mathbb{Q}(\alpha)$.

Obviously $\alpha\in \mathbb{Q}(\alpha)$ so $\sum q\alpha^i\in\mathbb{Q}(\alpha)$ for $q\in\mathbb{Q}$.

But what next, how do I approach these type of problems?

Answer 1

$$ a=\sqrt{5}+\sqrt[5]{2}\quad\Longrightarrow\quad(a-\sqrt{5})^5=2 $$ Hence $$ a^5-5a^4\sqrt{5}+50a^3-50a^2\sqrt{5}+250a-25\sqrt{5}=2 $$ and thus $$ \sqrt{5}=\frac{a^5+50a^3+250a-2}{5a^4+50a^2+25}\in\mathbb Q[a]. $$

Answer 2

Hint:

$\bigl(\alpha-\sqrt5\bigr)^5=2$.